%I #33 Mar 04 2022 14:19:46
%S 1,2,4,6,8,12,18,30,16,24,36,60,54,90,150,210,32,48,72,120,108,180,
%T 300,420,162,270,450,630,750,1050,1470,2310,64,96,144,240,216,360,600,
%U 840,324,540,900,1260,1500,2100,2940,4620,486,810,1350,1890,2250,3150,4410
%N a(0) = 1, and for n > 0, a(n) = A000040(A000120(n)) * a(floor(n/2)); essentially sequence A055932 generated using A000120, hence sorted by number of factors.
%C Note that for n>0 the prime divisors of a(n) are consecutive primes starting with 2. All of the least prime signatures (A025487) are included; with the other values forming A056808.
%C Using the formula, terms of b(n)= a(n)/A057334(n) are: 1, 1, 2, 2, 4, 4, 6, 6, 8, ..., indeed a(n) repeated. - _Michel Marcus_, Feb 09 2014
%C a(n) is the unique normal number whose unsorted prime signature is the k-th composition in standard order (graded reverse-lexicographic). This composition (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. A number is normal if its prime indices cover an initial interval of positive integers. Unsorted prime signature is the sequence of exponents in a number's prime factorization. - _Gus Wiseman_, Apr 19 2020
%H Michael De Vlieger, <a href="/A057335/b057335.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = A057334(n) * a (repeated).
%F A334032(a(n)) = n; a(A334032(n)) = A071364(n). - _Gus Wiseman_, Apr 19 2020
%F a(n) = A122111(A019565(n)); A019565(n) = A122111(a(n)). - _Peter Munn_, Jul 18 2020
%F a(n) = A336321(2^n). - _Peter Munn_, Mar 04 2022
%e From _Gus Wiseman_, Apr 19 2020: (Start)
%e The sequence of terms together with their prime indices begins:
%e 1: {}
%e 2: {1}
%e 4: {1,1}
%e 6: {1,2}
%e 8: {1,1,1}
%e 12: {1,1,2}
%e 18: {1,2,2}
%e 30: {1,2,3}
%e 16: {1,1,1,1}
%e 24: {1,1,1,2}
%e 36: {1,1,2,2}
%e 60: {1,1,2,3}
%e 54: {1,2,2,2}
%e 90: {1,2,2,3}
%e 150: {1,2,3,3}
%e 210: {1,2,3,4}
%e 32: {1,1,1,1,1}
%e 48: {1,1,1,1,2}
%e For example, the 27th composition in standard order is (1,2,1,1), and the normal number with prime signature (1,2,1,1) is 630 = 2*3*3*5*7, so a(27) = 630.
%e (End)
%t Table[Times @@ Map[If[# == 0, 1, Prime@ #] &, Accumulate@ IntegerDigits[n, 2]], {n, 0, 54}] (* _Michael De Vlieger_, May 23 2017 *)
%o (PARI) mg(n) = if (n==0, 1, prime(hammingweight(n))); \\ A057334
%o lista(nn) = {my(v = vector(nn)); v[1] = 1; for (i=2, nn, v[i] = mg(i-1)*v[(i+1)\2];); v;} \\ _Michel Marcus_, Feb 09 2014
%o (PARI) A057335(n) = if(0==n,1,prime(hammingweight(n))*A057335(n\2)); \\ _Antti Karttunen_, Jul 20 2020
%Y Cf. A000120, A057334, A055932 and A056808.
%Y Cf. A324939.
%Y Unsorted prime signature is A124010.
%Y Numbers whose prime signature is aperiodic are A329139.
%Y The reversed version is A334031.
%Y A partial inverse is A334032.
%Y All of the following pertain to compositions in standard order (A066099):
%Y - Length is A000120.
%Y - Sum is A070939.
%Y - Strict compositions are A233564.
%Y - Constant compositions are A272919.
%Y - Aperiodic compositions are A328594.
%Y - Normal compositions are A333217.
%Y - Permutations are A333218.
%Y - Heinz number is A333219.
%Y Cf. A029931, A048793, A052409, A056239, A066099, A112798, A124767, A228351, A233249, A333220.
%Y Related to A019565 via A122111 and to A000079 via A336321.
%K easy,nonn
%O 0,2
%A _Alford Arnold_, Aug 27 2000
%E More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003
%E New primary name from _Antti Karttunen_, Jul 20 2020